2022 AMC 10A Problems/Problem 15
Contents
Problem
Quadrilateral 
with side lengths 
is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form 
where 
and 
are positive integers such that 
and have no common prime factor. What is 
Diagram
![[asy] /* Made by MRENTHUSIASM */ size(200); pair O, A, B, C, D; O = origin; A = (-25/2,0); C = (25/2,0); B = intersectionpoints(Circle(A,7),Circle(C,24))[0]; D = intersectionpoints(Circle(A,15),Circle(C,20))[1]; fill(Circle(O,25/2),yellow); fill(A--B--C--D--cycle,white); dot("$A$",A,1.5*W,linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*E,linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot(O,linewidth(4)); draw(Circle(O,25/2)); draw(A--B--C--D--cycle); label("$7$",midpoint(A--B),rotate(90)*dir(midpoint(A--B)--A)); label("$24$",midpoint(B--C),rotate(-90)*dir(midpoint(B--C)--B)); label("$20$",midpoint(C--D),rotate(-90)*dir(midpoint(C--D)--C)); label("$15$",midpoint(D--A),rotate(90)*dir(midpoint(D--A)--D)); [/asy]](http://latex.artofproblemsolving.com/6/f/4/6f4d2e5edead8a1a9eddcc202b507b2a4f6e8023.png)
~MRENTHUSIASM
Solution 1 (Inscribed Angle Theorem)
Opposite angles of every cyclic quadrilateral are supplementary, so ![\[\angle B + \angle D = 180^{\circ}.\]](http://latex.artofproblemsolving.com/1/0/b/10b835ac4f345b1691b1f847eb0bd5445af3efd0.png)
We claim that 
We can prove it by contradiction:
- If
then 
and 
are both acute angles. This arrives at a contradiction.
then 
and 
are both acute angles. This arrives at a contradiction.- If
then and are both obtuse angles. This arrives at a contradiction.
then (This can also be shown using Law of Cosines: Since the sum of the squares of (7 and 24) and of (20 and 15) are equal, and the cosines of the corner angles are additive opposites, the cosines of the angles must 0. --pinava)
By the Inscribed Angle Theorem, we conclude that 
is the diameter of the circle. So, the radius of the circle is 
The area of the requested region is ![\[\pi r^2 - \frac12\cdot AB\cdot BC - \frac12\cdot AD\cdot DC = \frac{625\pi}{4}-\frac{168}{2}-\frac{300}{2}=\frac{625\pi-936}{4}.\]](http://latex.artofproblemsolving.com/e/a/b/eabc501f3d3709f5610b40fa8eef1209001f1b06.png)
Therefore, the answer is 
~MRENTHUSIASM
Solution 2 (Brahmagupta‘s Formula)
When we look at the side lengths of the quadrilateral we see 
and 
which screams out 
because of Pythagorean triplets. As a result, we can draw a line through points 
and 
to make a diameter of 
See Solution 1 for a rigorous proof.
Since the diameter is 
we can see the area of the circle is just 
from the formula of the area of the circle with just a diameter.
Then we can use Brahmagupta Formula 
where 
are side lengths, and 
is semi-perimeter to find the area of the quadrilateral.
If we just plug the values in, we get 
So now the area of the region we are trying to find is 
Therefore, the answer is
~Gdking
Solution 3 (Circumradius's Formula)
We can guess that this quadrilateral is actually made of two right triangles: 
has a 
ratio in the side lengths, and 
is a 
triangle.
(See Solution 1 for a proof.)
Next, we can choose one of these triangles and use the circumradius formula to find the radius. Let's choose the 
triangle. The area of the triangle is equal to the product of the side lengths divided by 
times the circumradius. Therefore, 
. Solving this simple algebraic equation gives us 
.
Plugging in the values, we have 
. Rewriting this gives us 
.
Therefore, adding these values gets us 
Video Solution 1
~Education, the Study of Everything
Video Solution 2
See Also
| 2022 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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